# Is there any identity which cannot be proved

For example, if we want to prove that $a^2+b^2\ge 2ab$ for all $a,b\in\mathbb{R}$, we will start from something which is true (axiom or something that is already proved). In this case we will use fact that square of any real number cannot be negative, so $(a-b)^2\ge0$. Transforming this inequality we will get $a^2+b^2\ge 2ab$. This is one of the simplest proof. We started from something which is aready proved and transforming it we got required inequality. But what about complex equations or inequalities? Why is so hard to prove that if $a,b,c,x,y,z\in\mathbb{N}$ and $x,y,z>2$ such that $a^x+b^y=c^z$ then $a,b,c$ must have a common prime divisor? My question: is there any equation, inequality or anything which never can be proved using axioms or identities which are already proved?

• Think in set theory, this equality cant have a proof with the axioms we have: $|\mathbb{R}|=\aleph_{1}$. This was a problem that Cantor tried to prove al his life, it has an interesting history. – Yesid Dec 25 '14 at 15:44
• – Git Gud Dec 25 '14 at 15:45
• Prove from what? Take anything as an axiom, it will prove itself. – Asaf Karagila Dec 25 '14 at 18:26
• Also, it should be noted that there has been several threads, all highly voted and thoroughly answered, about "unprovable statements". For example this and this. And probably many others too. – Asaf Karagila Dec 25 '14 at 18:32

• Exhibit a set $X$ such that $f:\mathbb N\to X$ is an injective function, but no function $F:X\to\mathbb N$ is injective, and $g:X\to\mathbb R$ is an injective function, but no function $G:\mathbb R\to X$ is injective. [Click for hint.]
• $P\neq NP$. If you can prove or disprove it, or at least prove that it's undecidable, you'll be very famous. And rich.
• @Mathematician171 Fair enough—the “$P=NP?$” problem may have a definitive answer, albeit it is apparently very difficult to find. But the continuum hypothesis is proven to be undecidable using the standard axioms of modern mathematics, the so-called Zermelo-Fraenkel model with the axiom of choice (ZFC). – triple_sec Dec 25 '14 at 15:48
• I can easily prove $\sf CH$. – Asaf Karagila Dec 25 '14 at 18:36
• With $V=L$ as an additional axiom. :-) – Asaf Karagila Dec 25 '14 at 18:49